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in Technique[技术] by (71.8m points)

javascript - I'm trying to use a simple webcam capture upload, upload isn't working?

UPDATE:

This is the entire code, which I pretty much copied and pasted.

<!DOCTYPE HTML>
<head>
<script type="text/javascript" src="http://code.jquery.com/jquery-latest.js"></script>
<script language="JavaScript" type="text/javascript"></script>

<style>

</style>
</head>
<body>
<script>
Webcam.set({
    width: 320,
    height: 240,
    dest_width: 640,
    dest_height: 480,
    image_format: 'jpeg',
    jpeg_quality: 90,
    force_flash: false
});
</script>
<div id="web_camera" style="width:320px; height:240px;"></div>
<div id="cam_result"></div>
 <script type="text/javascript" src="webcam.js"></script>
  <script language="JavaScript">
   document.addEventListener("DOMContentLoaded", function(event) {
        Webcam.set({
            width: 320,
            height: 240,
            image_format: 'jpeg',
            jpeg_quality: 90
        });
        Webcam.attach( '#web_camera' );
        function take_snapshot() {
            // take snapshot and get image data
            Webcam.snap( function(data_uri) {
                // display results in page
                document.getElementById('cam_results').innerHTML =
                    '<h2>Here is your image:</h2>' +
                    '<img src="'+data_uri+'"/>';
                                Webcam.upload( data_uri, 'upload.php', function(code, text) {
                                            // Upload complete!
                                            // 'code' will be the HTTP response code from the server, e.g. 200
                                            // 'text' will be the raw response content
                                });
            } );
        }
   });
  </script>
<a href="javascript:void(take_snapshot())">Take Snapshot</a>

</body>

I'm using this link

http://mycodingtricks.com/javascript/webcam-api/

This one looks a lot better but may be the same thing

http://www.html5rocks.com/en/tutorials/getusermedia/intro/

What I'm concerned about is the data_uri, also the url upload

So the webcam works, shows my face, whatever, but I push this

<a href="javascript:void(take_snapshot())">Take Snapshot</a>

and nothing happens. I see the little grey box at the bottom left saying javascript:void(take_snapshot()) I'm wondering if I'm supposed to put a parameter...

There can be several problems, I am using domain mapping and the folder may be pointed incorrectly or it could be a file permission problem, I did chown with www-data

This is the upload.php as suggested or rather given by the first link

<?php
    // be aware of file / directory permissions on your server
    move_uploaded_file($_FILES['webcam']['tmp_name'], '/tabdater/uploads/webcam'.md5(time()).rand(383,1000).'.jpg');
?>

I'd appreciate any help.

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1 Reply

0 votes
by (71.8m points)

This should be a great starting point for you. I hope this helps you.

<!doctype html>
<html lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>Cam Snap</title>
<script type="text/javascript" src="webcam.js"></script>
<script language="JavaScript">
function take_snapshot() {
    Webcam.snap(function(data_uri) {
    document.getElementById('results').innerHTML = '<img id="base64image" src="'+data_uri+'"/><button onclick="SaveSnap();">Save Snap</button>';
});
}
function ShowCam(){
Webcam.set({
width: 320,
height: 240,
image_format: 'jpeg',
jpeg_quality: 100
});
Webcam.attach('#my_camera');
}
function SaveSnap(){
    document.getElementById("loading").innerHTML="Saving, please wait...";
    var file =  document.getElementById("base64image").src;
    var formdata = new FormData();
    formdata.append("base64image", file);
    var ajax = new XMLHttpRequest();
    ajax.addEventListener("load", function(event) { uploadcomplete(event);}, false);
    ajax.open("POST", "upload.php");
    ajax.send(formdata);
}
function uploadcomplete(event){
    document.getElementById("loading").innerHTML="";
    var image_return=event.target.responseText;
    var showup=document.getElementById("uploaded").src=image_return;
}
window.onload= ShowCam;
</script>
<style type="text/css">
.container{display:inline-block;width:320px;}
#Cam{background:rgb(255,255,215);}#Prev{background:rgb(255,255,155);}#Saved{background:rgb(255,255,55);}
</style>
</head>
<body>
<div class="container" id="Cam"><b>Webcam Preview...</b>
    <div id="my_camera"></div><form><input type="button" value="Snap It" onClick="take_snapshot()"></form>
</div>
<div class="container" id="Prev">
    <b>Snap Preview...</b><div id="results"></div>
</div>
<div class="container" id="Saved">
    <b>Saved</b><span id="loading"></span><img id="uploaded" src=""/>
</div>
</body>
</html>

PHP (Must have the uploads directory) -

<?php
define('UPLOAD_DIR', 'uploads/');
$img = $_POST['base64image'];
$img = str_replace('data:image/jpeg;base64,', '', $img);
$img = str_replace(' ', '+', $img);
$data = base64_decode($img);
$file = UPLOAD_DIR . uniqid() . '.png';
$success = file_put_contents($file, $data);
print $success ? $file : 'Unable to save the file.';
?>

Credit to This Blog for the php!


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